\(\left(2x-1\right)^{2023}=\left(2x-1\right)^{2024}\) \((x\in\mathbb{N})\)
\(\Rightarrow\left(2x-1\right)^{2023}-\left(2x-1\right)^{2024}=0\)
\(\Rightarrow\left(2x-1\right)^{2023}\left[1-\left(2x-1\right)\right]=0\)
\(\Rightarrow\left(2x-1\right)^{2023}\left(1-2x+1\right)=0\)
\(\Rightarrow\left(2x-1\right)^{2023}\left(2-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=0\\2-2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
Vậy \(x=1\) là giá trị cần tìm.
\(\left(2x-1\right)^{2023}=\left(2x-1\right)^{2024}\)
=>\(\left(2x-1\right)^{2024}-\left(2x-1\right)^{2023}=0\)
=>\(\left(2x-1\right)^{2023}\left[\left(2x-1\right)-1\right]=0\)
=>\(\left(2x-1\right)^{2023}\cdot\left(2x-2\right)=0\)
=>\(\left[{}\begin{matrix}2x-1=0\\2x-2=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
